# Is $e$ a zero of power series with rational coefficients?

Since $$e$$ is a transcendence number, so it is certain that it is not zero of any polynomial with rational coefficients. However, I wonder can we find a power series with rational coefficient such that it is zero evaluated at $$e$$. If such series exists, can we explicitly find the coefficients?

• Ethan Bolker's answer is great. I just wanted to comment on the last line, about series that are particularly nice. You are perhaps thinking about $\pi$ that is the zero of a particularly nice power series: the Taylor series of the sin function. Now one thing that sets $\pi$ apart from $e$ despite both of them being transcendental is that $\pi$ is a 'period' and $e$ conjecturally is not. I am inclined to believe b/c of this that no 'particularly nice' series for $e$ exists. The answers to my earlier MO question about this topic: mathoverflow.net/q/180035/41139 might be helpful here. Nov 25, 2021 at 16:05
• This is, of course, all speculation Nov 25, 2021 at 16:08
• Similar to Ethan Bolker's answer you can just use reciprocals so something like $1-\frac{x}{3}-\frac{x^2}{79}-\frac{x^3}{53748}-\ldots$ and even those terms have got you to below $10^{-8}$ and the next to below $10^{-19}$ Nov 25, 2021 at 16:16

Start with $$a_0 = 1$$ (or anywhere else you like).
Now find a rational multiple $$a_1$$ of $$e$$ such that $$0 < 1 + a_1e < 1/2.$$ Then find rational $$a_2$$ such that $$0 < 1 + a_1e + a_2e^2< 1/4.$$ Continue in the obvious way.